1
Nothing cancels until everything is factored.
You may cancel a factor. You may never cancel a term.
not allowed
x² + 2x − 15 over x² − 25
x² is a term, joined by + and −
allowed
(x + 5)(x − 3) over (x + 5)(x − 5)
a whole bracket, top and bottom
The one test
Can you draw a bracket around it and see it multiplied by everything else on
that line? Then it is a factor and it can go. If a plus or a minus sign
separates it from its neighbour, it is a term and it stays.
The SAT never says “simplify”. It says “Which expression is equivalent to
the given expression?” — which means the same thing, and means the two
expressions must agree at every x that is allowed.
2
Factor top and bottom, then cancel the matching bracket.
Back to the opening problem. Two perfect squares with a minus between them always split; for the top, find the pair.
| Pair | Multiply to −15 | Add to 2 |
|---|
| −1 and 15 | −15 | 14 |
|---|
| −3 and 5 | −15 | 2 |
|---|
| 3 and −5 | −15 | −2 |
|---|
x² + 2x − 15 = (x + 5)(x − 3)
the pair −3 and 5
x² − 25 = (x + 5)(x − 5)
difference of squares
(x + 5) ÷ (x + 5) = 1
a whole bracket, so it goes
answer: x − 3x − 5
nothing left to cancel
x − 3x − 5
It will not go further. x − 3 and x − 5 share no factor, and the loose x’s are
terms. Check it if you like: at x = 1 the original is 1 + 2 − 151 − 25 =
−12−24 = 0.5, and 1 − 31 − 5 = −2−4 = 0.5 ✓
3
The excluded values come from the original denominator.
A bracket you cancelled still counts — it was down there when the question was written.
x + 5 = 0 → x = −5
the bracket you cancelled
x − 5 = 0 → x = 5
still on the bottom
so x ≠ 5 and x ≠ −5
exactly the line under the stem
Cancelling hides a hole; it does not fill it
The simplified x − 3x − 5 looks perfectly happy at x = −5, where it gives
−8/−10 = 0.8. The original is 0 ÷ 0 there — undefined. That is why the SAT
prints the “where x ≠ …” line for you: it is telling you which values were
already thrown out, so the two expressions really are equivalent everywhere
that is left.
Read that line the other way round and it is a hint. “where x ≠ 5 and x ≠ −5”
is announcing that the denominator factors as (x + 5)(x − 5) before you have
factored a thing.
4
Multiplying is easy; dividing is multiplying after one flip.
Factor everything, then cancel any bracket on a top against the same bracket on a bottom — across the multiplication sign is fine.
x² − 4x² + 7x + 12 ÷
x − 2x + 3
= (x − 2)x + 2(x + 3)(x + 4) · x + 3x − 2
factor, then flip the divisor
(x − 2) cancels, (x + 3) cancels
top against bottom, either fraction
= x + 2x + 4
what is left
x + 2x + 4, with x ≠ −3, −4 and 2
Flip the second fraction only — never the first, and never both. And the
exclusions gather from three places: the two starting denominators and the
numerator of the divisor, because you moved it downstairs.
5
To add or subtract, both denominators must match first.
Exactly as with 13 + 15 — you cannot add until both are in fifteenths. Factor each denominator, take every factor once, and give each numerator what it is missing.
4x + 2 +
3x − 6
LCD = (x + 2)(x − 6)
no factor in common, so both
= [4(x − 6) + 3(x + 2)]/((x + 2)(x − 6))
each numerator gains the other bracket
4x − 24 + 3x + 6 = 7x − 18
expand and collect the top
= 7x − 18(x + 2)(x − 6)
nothing cancels; x ≠ −2, 6
Never add the denominators
4x + 2 + 3x − 6 is not 72x − 4. It is the same instinct that would
turn 13 + 15 into 28, and it is wrong in both places. Leave the finished
denominator factored, too — it is easier to spot a cancellation, and the
excluded values are sitting there in plain sight.
When the denominators do share a factor, the LCD is smaller than you
feared. For 2xx² − 9 and 1x + 3: factor the first bottom into
(x − 3)(x + 3) and the second bottom is already inside it, so the LCD is just
(x − 3)(x + 3). Multiplying the two together would give (x − 3)(x + 3)² —
correct, but twice the work and it needs cancelling at the end.
6
Subtracting: the minus sign hits the whole numerator.
Bracket the numerator you are subtracting before you touch it.
2xx² − 9 −
1x + 3
= 2x(x − 3)(x + 3) − x − 3(x − 3)(x + 3)
the second gains (x − 3)
top = 2x − (x − 3)
brackets first, always
= 2x − x + 3 = x + 3
both signs flip
= x + 3(x − 3)(x + 3) = 1x − 3
and now it cancels
1x − 3, with x ≠ 3 and x ≠ −3
“2x − x − 3” is the classic wrong line
Drop the bracket and you lose the sign on the 3, the top becomes x − 3, and
the fraction stubbornly refuses to simplify. If your answer will not cancel
when it looks like it should, check that subtraction first. A related trick
worth having: x − 3 and 3 − x differ only by a sign, so 13 − x =
−1x − 3. Pulling out a −1 makes two denominators match that looked
different.
7
On test day, one number settles an “equivalent expression” item.
Equivalent means equal at every allowed x — so a single legal x kills every choice but one, nearly every time.
pick x
x = 1
avoid
x = 0, ±5
| At x = 1 | Value | Verdict |
|---|
| given expression | −12−24 = 0.5 | the target |
|---|
| A. x + 3x + 5 | 46 ≈ 0.667 | out |
|---|
| B. x − 3x − 5 | −2−4 = 0.5 | survives |
|---|
| C. 2x − 15−25 | −13−25 = 0.52 | out |
|---|
| D. x + 5x − 5 | 6−4 = −1.5 | out |
|---|
Desmos is on your screen — use it
The Digital SAT gives you a graphing calculator the whole time. Type the given
expression as y₁ and a choice as y₂; if the two curves lie on top of each
other, that choice is the answer. Or open the table and compare one column of
values. Factor by hand when it is quick — it usually is — and let the
calculator confirm.
Avoid x = 0 when you test-substitute: it collapses too much and often makes two
wrong choices agree with the right one. Avoid any excluded value too — the
given expression is not defined there, so there is nothing to compare against.
✓
Check: did the factoring survive a test number?
original
1 + 2 − 151 − 25
= −12/−24 = 0.5
answer
1 − 31 − 5
= −2/−4 = 0.5 ✓
bans
x ≠ 5, x ≠ −5
from the original bottom ✓
Remember the flow
Factor · Cancel brackets · Combine over one denominator · State the bans
Terms never cancel — only factors.
